a: \(6^{x}:3^3:4=2\)

=>\(6^{x}=2\cdot4\cdot3^3=2^3\cdot3^3=6^3\)

=>x=3

b: \(3\left(x-5\right)^3=81\)

=>\(\left(x-5\right)^3=\frac{81}{3}=27=3^3\)

=>x-5=3

=>x=3+5=8

c: \(7x-2x=6^{17}:6^{15}+3^8:2^6\)

=>\(5x=6^2+\frac{3^8}{2^6}=36+\frac{6561}{64}=\frac{8865}{64}\)

=>\(x=\frac{8865}{64}:5=\frac{1773}{64}\)

Bài 3: (d) cắt (d2) tại điểm C nằm trên trục tung

=>\(\begin{cases}a<>a^{\prime}\\ b=b^{\prime}\end{cases}\Rightarrow\begin{cases}m+3<>m-4\\ 2m-10=-2m-8\end{cases}\Rightarrow2m-10=-2m-8\)

=>2m+2m=-8+10

=>4m=2

=>m=0,5

Khi m=0,5 thì(d): y=(0,5+3)x+2*0,5-10=3,5x-9

Khi m=0,5 thì (d2): y=(0,5-4)x-2*0,5-8=-3,5x-9

Tọa độ A là:

\(\begin{cases}y=0\\ 3,5x-9=0\end{cases}\Rightarrow\begin{cases}y=0\\ 3,5x=9\end{cases}=>\begin{cases}y=0\\ x=\frac{9}{3,5}=\frac{18}{7}\end{cases}\)

Tọa độ B là:

\(\begin{cases}y=0\\ -3,5x-9=0\end{cases}\Rightarrow\begin{cases}y=0\\ -3,5x=9\end{cases}\Rightarrow\begin{cases}y=0\\ x=-\frac{18}{7}\end{cases}\)

B(-18/7;0); C(0;-9); A(18/7;0)

\(BC=\sqrt{\left(0+\frac{18}{7}\right)^2+\left(-9-0\right)^2}=\sqrt{\left(\frac{18}{7}\right)^2+9^2}=\sqrt{\frac{324}{49}+81}=\sqrt{\frac{4293}{49}}=\frac{\sqrt{4293}}{7}\)

\(AC=\sqrt{\left(\frac{18}{7}-0\right)^2+\left(0+9\right)^2}=\sqrt{\left(\frac{18}{7}\right)^2+9^2}=\frac{\sqrt{4293}}{7}\)

\(OA=\sqrt{\left(\frac{18}{7}-0\right)^2+\left(0-0\right)^2}=\frac{18}{7}\)

\(OB=\sqrt{\left(-\frac{18}{7}-0\right)^2+\left(0-0\right)^2}=\frac{18}{7}\)

Vì OA=OB và AC=BC

nên \(\frac{OA}{BC}=\frac{OB}{AC}\)

Câu 4:

Xét ΔOAI và ΔOBI có

OA=OB

AI=BI

OI chung

Do đó: ΔOAI=ΔOBI

`a)5y(2y-1)-(3y+2)(3-3y)`

`=10y^2-5y+(3y+2)(3y-3)`

`=10y^2-5y+9y^2-9y+6y-6`

`=19y^2-8y-6`

`b)(6x+1)^2-2(6x+1)(6x-1)+(6x-1)^2`

`=(6x+1-6x+1)^2`

`=2^2=4`

`c)(2x+3)^2-2(2x+3)(x-20+(x-2)^2`

`=(2x+3-x+2)^2`

`=(x+5)^2`

`=x^2+10x+25`

\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)

\(\text{a) }\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\\ =\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\\ =\left[6x+1-6x+1\right]^2\\ =2^2\\ =4\)

\(\text{b) }3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1\)

\(1,\left(x-2\right)\left(x+2\right)\left(x^2+4\right)-\left(x^2-3\right)\left(x^2+3\right)\)

\(=\left(x^2-4\right)\left(x^2+4\right)-\left(x^2-9\right)\)

\(=x^2-16-x^2+9\)

\(=-7\)

\(2,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1-6x+1\right)^2\)

\(=2^2=4\)

\(27+\left(220-6x\right).3=615\)

\(\rightarrow\left(220-6x\right).3=615-27=588\)

\(\rightarrow220-6x=588:3=196\)

\(\rightarrow6x=220-196=24\)

\(\rightarrow x=24:6=4\)

\(315-\left(x-3\right).7=210\)

\(\Rightarrow\left(x-3\right).7=315-210=105\)

\(\Rightarrow x-3=105:7=15\)

\(\Rightarrow x=15+3=18\)

a: \(6x^2+13x-5\)

\(=6x^2+15x-2x-5\)

=3x(2x+5)-(2x+5)

=(2x+5)(3x-1)

b: \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1-6x+1\right)^2=2^2=4\)

c: \(\left(x-1\right)^2\ge0\forall x\)

=>\(x^2-2x+1\ge0\forall x\)

=>\(x^2-2x+1+2\ge0+2\forall x\)

=>\(x^2-2x+3\ge2\forall x\)